5r^2+6r+1=0

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Solution for 5r^2+6r+1=0 equation:



5r^2+6r+1=0
a = 5; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·5·1
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4}{2*5}=\frac{-10}{10} =-1 $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4}{2*5}=\frac{-2}{10} =-1/5 $

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